2025 Cambridge IGCSE Science - Combined (0653) Practice Paper with Answers

At temperatures significantly above the optimum, the increased kinetic energy causes the atoms in the enzyme molecule to vibrate so violently that the bonds holding its specific three-dimensional shape together are broken. This alters the shape of the active site permanently (denaturation), so the substrate can no longer fit into it. Enzymes are proteins (not glucose polymers), and their kinetic energy increases rather than decreases with temperature, making option B the correct explanation.

1 mark for identifying that denaturation involves a permanent change in the shape of the active site, preventing substrate binding.

Benedict's test is used to detect reducing sugars; a blue result is negative, indicating that reducing sugars are absent. The Biuret test is used to detect proteins; a purple (or violet) color is positive, indicating that protein is present. Therefore, the sample contains protein but no reducing sugar.

1 mark for identifying that a negative Benedict's test (remains blue) indicates the absence of reducing sugars and a positive Biuret test (turns purple) indicates the presence of protein.

In Group I (alkali metals), the melting points decrease as the proton number increases. Reactivity increases down Group I but decreases down Group VII. For Group VII, physical state changes from gas to liquid to solid as proton number increases. Thus, option C is the correct trend.

1 mark for identifying that Group I melting points decrease as the proton number increases.

In the blast furnace, carbon monoxide (\text{CO}) reacts with iron(III) oxide (\text{Fe}_2\text{O}_3) to reduce it to molten iron, while being oxidized to carbon dioxide (\text{CO}_2). The chemical equation is: \text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2. Therefore, carbon monoxide is the primary reducing agent.

1 mark for identifying carbon monoxide as the primary reducing agent.

Using the formula for final velocity under constant acceleration: \(v = u + at\), where \(u = 0\text{ m/s}\) (starts from rest), we get: \(v = 0 + (2.5 \times 6.0) = 15\text{ m/s}\). The distance travelled \(d\) can be calculated using \(d = ut + \frac{1}{2}at^2\) or the average speed method: \(d = \text{average speed} \times t = \frac{0 + 15}{2} \times 6.0 = 7.5 \times 6.0 = 45\text{ m}\). Therefore, option A is correct.

1 mark for correctly calculating both the final speed of 15 m/s and the distance of 45 m.

When resistors are in series, their resistances add up: \(R_{\text{series}} = R + R = 2R\). When identical resistors are in parallel, their combined resistance is given by: \(1/R_{\text{parallel}} = 1/R + 1/R = 2/R\), so \(R_{\text{parallel}} = R/2 = 0.5R\). Therefore, option A is correct.

1 mark for correctly determining the combined series resistance as 2R and the combined parallel resistance as 0.5R.

All electromagnetic waves travel at the same speed of \(3 \times 10^8\text{ m/s}\) in a vacuum. Ultraviolet waves have a higher frequency than infrared waves. Sound waves are longitudinal mechanical waves, and radio waves (which are electromagnetic) can easily travel through a vacuum. Therefore, option B is correct.

1 mark for identifying that all electromagnetic waves travel at the same speed in a vacuum.

Alkenes are unsaturated hydrocarbons containing a carbon-carbon double bond (\text{C}=\text{C}). They undergo an addition reaction with aqueous bromine, where bromine atoms add across the double bond, which rapidly turns the orange-brown solution colorless. Alkanes are saturated and do not react with bromine water unless UV light is present, in which case a substitution reaction occurs slowly. Hence, the rapid decolorization indicates an alkene and an addition reaction.

1 mark for identifying the hydrocarbon as an alkene and the reaction type as an addition reaction.

Both typical plant cells and animal cells contain a cell membrane, cytoplasm, and a nucleus. Only plant cells contain a cellulose cell wall, chloroplasts, and a large permanent vacuole. Animal cells do not have a cell wall or chloroplasts, and their vacuoles (if present) are small and temporary.

1 mark for option B. 0 marks for other choices.

Isotopes are atoms of the same element that have the same proton number but a different nucleon number (due to a different number of neutrons). The original phosphorus atom has 15 protons and \(31 - 15 = 16\) neutrons. A different isotope of phosphorus must still have 15 protons, but a different number of neutrons. Option A has 15 protons and 17 neutrons, which represents a different isotope. Options C and D represent atoms of a different element entirely, since their proton number is 16.

1 mark for option A. 0 marks for other choices.

The total distance travelled is equal to the area under the speed-time graph. For the first stage (constant acceleration from rest to \(12\text{ m/s}\) in \(4.0\text{ s}\)), the distance is the area of a triangle: \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = 0.5 \times 4.0\text{ s} \times 12\text{ m/s} = 24\text{ m}\). For the second stage (constant speed of \(12\text{ m/s}\) for \(6.0\text{ s}\)), the distance is the area of a rectangle: \(d_2 = \text{base} \times \text{height} = 6.0\text{ s} \times 12\text{ m/s} = 72\text{ m}\). The total distance is \(d_1 + d_2 = 24\text{ m} + 72\text{ m} = 96\text{ m}\).

1 mark for option C. 0 marks for other choices.

At low temperatures, such as \(5^\circ\text{C}\), enzymes and substrate molecules have very low kinetic energy, leading to a low frequency of successful collisions and thus a very low reaction rate. At high temperatures, such as \(80^\circ\text{C}\), enzymes are denatured because the high temperature disrupts the bonds maintaining their shape, changing the shape of the active site so that the substrate can no longer fit.

1 mark for option D. 0 marks for other choices.

Saturated hydrocarbons (alkanes) do not react with aqueous bromine, so the mixture remains orange-brown. Unsaturated hydrocarbons (alkenes) react with aqueous bromine and decolourise it. Thus, compound X must be an alkane, which eliminates ethene and hexene. Of the remaining alkanes, ethane is a gas at room temperature, while hexane is a liquid. Therefore, compound X must be hexane.

1 mark for option C. 0 marks for other choices.

In the electromagnetic spectrum, ordered by increasing wavelength (and decreasing frequency), the sequence of these three regions is: ultraviolet (shortest wavelength), followed by visible light, followed by infrared (longest wavelength). Therefore, the correct order of shortest to longest wavelength is 2 \(\rightarrow\) 3 \(\rightarrow\) 1.

1 mark for option C. 0 marks for other choices.

Osmosis is defined as the net movement of water molecules from a region of higher water potential (a dilute solution) to a region of lower water potential (a concentrated solution) through a partially permeable membrane.

1 mark for option B. 0 marks for other choices.

In a series circuit, the total resistance is the sum of the individual resistances: \(R_{\text{total}} = R_1 + R_2 = 2.0\ \Omega + 4.0\ \Omega = 6.0\ \Omega\). The total current from the battery is found using Ohm's law: \(I = \frac{V}{R_{\text{total}}} = \frac{6.0\text{ V}}{6.0\ \Omega} = 1.0\text{ A}\). In a series circuit, the current is identical at every point in the circuit. Therefore, the current through both \(R_1\) and \(R_2\) is \(1.0\text{ A}\).

1 mark for option A. 0 marks for other choices.

As Group I is descended from lithium to potassium, the melting points of the metals decrease and their reactivity with water increases.

1 mark for the correct option C.

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Inspired air contains approximately 21% oxygen and 0.04% carbon dioxide. Expired air contains less oxygen (around 16%) because some is absorbed into the blood, and more carbon dioxide (around 4%) because it is a waste product of respiration. Thus, inspired air contains more oxygen and less carbon dioxide than expired air.

1 mark for identifying that inspired air has more oxygen and less carbon dioxide than expired air.

The atomic number (11) is the number of protons. The mass number (23) is the total number of protons and neutrons. The number of neutrons is 23 - 11 = 12. Therefore, the nucleus contains 11 protons and 12 neutrons.

1 mark for the correct proton and neutron count.

At high temperatures, the thermal energy causes the enzyme molecule to vibrate excessively, breaking the weak bonds that hold its shape. This denatures the enzyme, changing the shape of the active site so that the substrate can no longer fit.

1 mark for explaining that high temperature causes the active site of the enzyme to change shape, preventing the substrate from fitting.

A dull, dark (black) surface is both the best absorber and the best emitter of infrared (thermal) radiation. Shiny or light-colored surfaces are poor absorbers and poor emitters.

1 mark for identifying a dull black surface as both the best absorber and best emitter.

Alkenes are unsaturated hydrocarbons containing double carbon-carbon bonds. They react with aqueous bromine in an addition reaction, which decolourises the bromine water (turning it from orange to colorless). Alkanes do not react with bromine water under normal conditions.

1 mark for choosing the correct reagent (bromine water) and correct color change (orange to colorless) for the alkene.

Charge \(Q\) is calculated using the formula \(Q = I \times t\), where \(I\) is the current in amperes and \(t\) is the time in seconds. Substituting the given values: \(Q = 0.5\text{ A} \times 60\text{ s} = 30\text{ C}\).

1 mark for calculating the charge using \(Q = I \times t\) to obtain 30 C.

Sound waves require a medium and travel as longitudinal waves (oscillations parallel to wave propagation). Light waves do not require a medium and travel as transverse waves (oscillations perpendicular to wave propagation).

1 mark for identifying sound waves as longitudinal and light waves as transverse.

Diluting the hydrochloric acid by adding water decreases the concentration of acid particles. A lower concentration means there are fewer reacting particles per unit volume, which reduces the frequency of collisions, thereby decreasing the rate of reaction.

1 mark for identifying that diluting the acid decreases the rate of reaction.

Enzymes are proteins that function as biological catalysts. They are not carbohydrates, they are denatured (not killed) by high temperatures, and their activity is highly affected by pH.

1 mark: correct option B.

The proton number is 11, so there are 11 protons. The nucleon number is 23, so the number of neutrons is 23 - 11 = 12. A positive ion \(X^+\) has lost one electron, so it has 11 - 1 = 10 electrons.

1 mark: correct option A.

The distance travelled is the area under the speed-time graph. For the first 4.0 s: \(\text{distance} = 0.5 \times 4.0 \times 12 = 24\text{ m}\). For the next 6.0 s: \(\text{distance} = 6.0 \times 12 = 72\text{ m}\). Total distance = \(24 + 72 = 96\text{ m}\).

1 mark: correct option B.

The black paper blocks light from reaching the covered part of the leaf. Since light is required for photosynthesis, no starch is produced in this region. The iodine solution remains brown (the negative test result for starch).

1 mark: correct option B.

Alkenes are unsaturated hydrocarbons containing C=C double bonds, which react rapidly with aqueous bromine (decolourising it). Alkanes are saturated (only C-C single bonds) and do not react with aqueous bromine in the absence of light.

1 mark: correct option B.

All electromagnetic waves are transverse waves, do not require a medium to travel (can travel through a vacuum), and travel at the same speed of \(3.0 \times 10^8\text{ m/s}\) in a vacuum. They have different wavelengths and frequencies.

1 mark: correct option A.

During gas exchange in the alveoli, oxygen is absorbed into the blood and carbon dioxide is excreted. Therefore, inspired air has more oxygen (approx 21% vs 16%) and less carbon dioxide (approx 0.04% vs 4%) than expired air.

1 mark: correct option A.

To measure the current flowing through a component, an ammeter must be connected in series with it. To measure the potential difference across the component, a voltmeter must be connected in parallel across it.

1 mark: correct option C.

First, calculate the combined resistance of the two 10 ohm resistors in parallel: 1 / Rp = 1 / 10 + 1 / 10 = 2 / 10, which gives Rp = 5.0 ohms. Next, calculate the total resistance of the series combination: Rtotal = Rp + 5.0 ohms = 5.0 + 5.0 = 10.0 ohms. Finally, use Ohm's law to find the current: I = V / Rtotal = 6.0 V / 10.0 ohms = 0.60 A.

1 mark for the correct answer C. 0.60 A. Reject other options based on incorrect calculations of parallel or series resistances.

At high temperatures, the thermal energy disrupts the bonds maintaining the tertiary structure of the enzyme. This causes the active site of the enzyme to change shape permanently (denature), meaning the substrate can no longer fit into it to form an enzyme-substrate complex.

1 mark for the correct answer B. Other options incorrectly describe the effect of temperature on the structure or kinetic energy.

The lower number (atomic number) is 15, which is the number of protons. The upper number (mass number) is 31, so the number of neutrons is 31 - 15 = 16. The ion has a 3- charge, meaning it has 3 more electrons than protons, so electrons = 15 + 3 = 18.

1 mark for the correct option B. Distractors represent common errors in counting atomic components or handling ionic charge.

Using the equation speed = distance / time, we can rearrange to find time = distance / speed. Time = (1.5 x 10^{11} m) / (3.0 x 10^8 m/s) = 500 s.

1 mark for correct calculation and choosing option B.

Benedict's solution remains blue when reducing sugars are absent. Iodine solution turns blue-black in the presence of starch. Biuret reagent turns purple in the presence of protein. Therefore, starch and protein are present.

1 mark for selecting B. Distractors represent incorrect interpretations of food test color changes.

Fe2O3 loses oxygen to form Fe, so it is reduced. CO gains oxygen to form CO2, meaning it is oxidized and acts as the reducing agent by removing oxygen from Fe2O3.

1 mark for identification of Fe2O3 as the reduced substance and CO as the reducing agent.

Using the wave equation: v = f * lambda, we can rearrange to find lambda = v / f. Lambda = (3.0 x 10^8 m/s) / (9.0 x 10^7 Hz) = 3.33 m.

1 mark for the correct calculation and selecting B.

Transpiration is the evaporation of water from the leaves of a plant. The rate of transpiration increases with higher temperature (increases kinetic energy of water molecules), lower humidity (creates a steeper concentration gradient of water vapor), and higher wind speed (removes water vapor from the leaf surface, maintaining the gradient).

1 mark for identifying the correct combination of low humidity, high temperature, and high wind speed as C.

The time between the 1st clap and the 21st clap represents exactly 20 intervals. In each interval, the sound travels to the wall and back, which is a total distance of \(2d\). Therefore, the total distance travelled by the sound in time \(t\) is \(20 \times 2d = 40d\). Using the relation \(\text{speed} = \frac{\text{distance}}{\text{time}}\), we get \(v = \frac{40d}{t}\).

1 mark for the correct formula C. Deduct or reject options that use the incorrect number of intervals (e.g. 21 intervals) or do not account for the round-trip distance of the sound wave.

At temperatures significantly above the optimum, the active site of the enzyme changes shape permanently due to the breaking of bonds holding the protein's tertiary structure. This prevents substrate molecules from binding to the active site, meaning the enzyme is denatured.

1 mark for identifying that denaturation involves the active site changing shape so it can no longer bind the substrate. Reject options suggesting low kinetic energy or complete breakdown into amino acids.

Catalytic cracking is a thermal decomposition reaction that breaks down long-chain, less useful alkanes into shorter, more useful alkanes and alkenes. It requires high temperatures and a catalyst (such as silicon dioxide or aluminium oxide).

1 mark for selecting B. Reject other options which describe addition reactions, combustion reactions, or hydrogenation.

For the parallel combination of two identical resistors of resistance \(R\), the equivalent resistance \(R_p\) is given by: \(\frac{1}{R_p} = \frac{1}{R} + \frac{1}{R} = \frac{2}{R}\), which gives \(R_p = \frac{R}{2} = 0.5R\). Adding this parallel combination in series with the third resistor of resistance \(R\) gives a total equivalent resistance of: \(R_{\text{total}} = R_p + R = 0.5R + R = 1.5R = \frac{3}{2}R\).

1 mark for the correct calculations of the parallel portion followed by adding the series resistor, leading to option C.

A 3+ charge indicates that the ion has lost 3 electrons relative to the neutral atom. Therefore, the neutral atom has \(10 + 3 = 13\) electrons, meaning its atomic (proton) number is 13. Since the nucleon number (protons + neutrons) is 27, the number of neutrons is \(27 - 13 = 14\).

1 mark for option A, indicating 13 protons and 14 neutrons. Reject other configurations that do not yield a net 3+ charge with 10 electrons or do not sum to 27 nucleons.

The pulmonary vein is responsible for carrying oxygenated blood (high oxygen concentration) from the lungs back to the left atrium of the heart so that it can be pumped around the body.

1 mark for identifying the correct combination of high oxygen concentration and flow from the lungs to the heart (Option B).

By the principle of conservation of energy, the gravitational potential energy lost is converted entirely into kinetic energy (since air resistance is negligible). Thus, \(E_k = \Delta E_p = mgh = 4.0 \text{ kg} \times 9.8 \text{ m/s}^2 \times 5.0 \text{ m} = 196 \text{ J}\).

1 mark for the correct application of \(mgh\) to find the kinetic energy, leading to option C.

In this reaction, copper oxide (\(\text{CuO}\)) loses oxygen to become copper (\(\text{Cu}\)), meaning it is reduced. Hydrogen (\(\text{H}_2\)) gains oxygen to become water (\(\text{H}_2\text{O}\)), meaning it is oxidized and acts as the reducing agent.

1 mark for identifying that copper oxide is reduced because it loses oxygen. Reject other choices that incorrectly identify what is oxidized, reduced, or the oxidizing agent.

Expired air contains about 4% carbon dioxide, whereas inspired air contains only about 0.04% carbon dioxide. Therefore, carbon dioxide is much more concentrated in expired air than in inspired air. Since carbon dioxide reacts with limewater to turn it cloudy, Flask Y (expired air) will turn cloudy significantly faster than Flask X.

1 mark: Correct identification of Flask Y and explanation of the higher carbon dioxide concentration in expired air.

As Group 1 is descended, the single outer shell electron is further from the nucleus and more shielded by inner shells. This makes it easier to lose, which increases reactivity. Simultaneously, the metallic bonding becomes weaker as the ionic radius increases, which causes the melting points to decrease.

1 mark: Correctly identifying both the increase in reactivity and the decrease in melting point down Group 1.

First, calculate the acceleration using Newton's second law: \(a = \frac{F}{m} = \frac{2.0\text{ N}}{0.5\text{ kg}} = 4.0\text{ m/s}^2\). Next, calculate the final velocity using the equation of motion: \(v = u + at = 0 + (4.0\text{ m/s}^2 \times 3.0\text{ s}) = 12\text{ m/s}\). Finally, calculate the kinetic energy: \(E_k = \frac{1}{2}mv^2 = \frac{1}{2} \times 0.5\text{ kg} \times (12\text{ m/s})^2 = 0.25 \times 144 = 36\text{ J}\).

1 mark: Correct calculation of the kinetic energy (36 J) using appropriate formulas for acceleration, velocity, and kinetic energy.

An increase in temperature increases the average kinetic energy of the particles. Consequently, they move faster and collide more frequently. Crucially, a much larger fraction of the colliding particles now possess kinetic energy equal to or greater than the activation energy, leading to more successful collisions per unit time.

1 mark: Correctly linking temperature rise to both increased frequency of collisions and a higher proportion of particles with energy equal to or greater than the activation energy.

Using Snell's Law for light passing from glass to air: \(n = \frac{\sin r}{\sin i}\), where \(i = 30^\circ\) is the angle in the glass and \(r\) is the angle of refraction in air. Rearranging this gives: \(\sin r = n \times
\sin i = 1.5 \times \sin 30^\circ = 1.5 \times 0.5 = 0.75\). Taking the inverse sine of 0.75 gives \(r \approx 48.6^\circ\).

1 mark: Correct application of Snell's Law and calculation of the angle of refraction as 48.6 degrees.

Ethene is an alkene containing a carbon-carbon double bond (\(\text{C}=\text{C}\)). It undergoes an addition reaction with aqueous bromine, rapidly decolorizing the orange-brown solution in the dark. Alkanes like ethane and methane only react with bromine in the presence of ultraviolet light via a substitution reaction.

1 mark: Correctly identifying ethene as the compound that reacts rapidly with aqueous bromine in the dark.

First, calculate the effective resistance of the parallel combination: \(1/R_p = 1/4.0 + 1/6.0 = 5/12 \implies R_p = 2.4\ \Omega\). Next, the total resistance of the series circuit is the sum of the parallel combination and the series resistor: \(R_{\text{total}} = R_p + 2.6\ \Omega = 2.4 + 2.6 = 5.0\ \Omega\). The total current delivered by the battery is \(I = V/R_{\text{total}} = 12\text{ V} / 5.0\ \Omega = 2.4\text{ A}\). Since the \(2.6\ \Omega\) resistor is in series with the combination, the total current of \(2.4\text{ A}\) flows directly through it.

1 mark: Correctly calculated parallel resistance, total resistance, and current of 2.4 A.

Palisade mesophyll cells are plant cells, which contain a cell wall (composed of cellulose), chloroplasts (for photosynthesis), and a large permanent vacuole. Human liver cells are animal cells and lack these three structures, although they share other structures like the cell membrane, cytoplasm, and nucleus.

1 mark: Correctly identifying the three organelles/structures found in palisade mesophyll cells but not in human liver cells.

Above the optimum temperature, the high thermal energy causes the bonds maintaining the three-dimensional structure of the enzyme's active site to break. This changes the shape of the active site (denaturation), so that the substrate can no longer fit, leading to a rapid decrease in the rate of reaction.

Award 1 mark for selecting the option explaining denaturation as the change in the active site shape preventing substrate binding.

Zinc powder has a larger surface area than zinc lumps of the same mass. A larger surface area increases the frequency of collisions between reactant particles (zinc and hydrogen ions), which increases the rate of reaction. The total volume of hydrogen gas produced remains the same because the mass of zinc is the same (and acid is in excess). Surface area does not change the activation energy.

Award 1 mark for identifying that zinc powder has a higher rate due to increased collision frequency per unit time.

All electromagnetic waves travel at the same speed in a vacuum (\(3 \times 10^8\text{ m/s}\)). In the electromagnetic spectrum, radio waves have the longest wavelengths (and lowest frequencies), while X-rays have very short wavelengths (and high frequencies). Therefore, radio waves have a longer wavelength than X-rays and travel at the same speed in a vacuum.

Award 1 mark for recognizing that radio waves have longer wavelengths than X-rays and that all EM waves travel at the same speed in a vacuum.

First, calculate the work done: \(\text{Work done} = \text{Force} \times \text{distance} = 400\text{ N} \times 12\text{ m} = 4800\text{ J}\). Next, calculate the useful power output: \(\text{Power} = \frac{\text{Work done}}{\text{time}} = \frac{4800\text{ J}}{6.0\text{ s}} = 800\text{ W}\).

Award 1 mark for showing correct formula use and calculating power as 800 W.

Use the formula \(I = \frac{Q}{t}\), where \(Q\) is the charge in coulombs and \(t\) is the time in seconds. First, convert time to seconds: \(t = 1.2 \times 60 = 72\text{ s}\). Then, calculate current: \(I = \frac{18\text{ C}}{72\text{ s}} = 0.25\text{ A}\).

Award 1 mark for converting time to seconds and dividing charge by time to get 0.25 A.

Ethene contains a carbon-carbon double bond, making it an unsaturated hydrocarbon. It can undergo addition polymerisation to form poly(ethene). Ethane is saturated and does not react with steam to form ethanol. Complete combustion of ethane in excess oxygen produces carbon dioxide and water, not carbon monoxide.

Award 1 mark for identifying ethene as unsaturated and capable of polymerisation to form poly(ethene).

Arteries carry blood away from the heart at high pressure; they have thick walls containing elastic fibres and muscle, and a relatively narrow lumen to maintain pressure. Veins carry blood back to the heart at low pressure, have thin walls, a wide lumen, and contain valves to prevent backflow.

Award 1 mark for identifying the correct features and pressure of an artery.

Greenhouse gases allow shorter-wavelength radiation from the Sun to pass through the atmosphere to warm the Earth's surface. The warmed surface then emits longer-wavelength thermal (infrared) radiation. Greenhouse gases absorb this infrared radiation and re-emit it in all directions, including back towards the Earth's surface, trapping heat.

Award 1 mark for selecting the statement that explains the absorption and re-emission of thermal infrared radiation by greenhouse gases.

Above the optimum temperature of 40 °C, the high thermal energy causes the bonds within the enzyme's globular structure to break. This alters the shape of the active site, meaning the substrate is no longer complementary to the active site and cannot bind. This is called denaturation.

1 mark: Identify that above 40 °C, the enzyme denatures because its active site changes shape, preventing substrate binding.

Hydrocarbon X decolourises aqueous bromine immediately, indicating it is unsaturated and contains a double bond (an alkene, like ethene). Hydrocarbon Y requires UV light to react with bromine via substitution, which is characteristic of saturated hydrocarbons (alkanes, like ethane). Ethanol is not a hydrocarbon.

1 mark: Correctly identify X as an alkene (ethene) and Y as an alkane (ethane) based on their reactions with aqueous bromine.

First, convert the wavelength from centimetres to metres: \(\lambda = 8.0\text{ cm} = 0.080\text{ m}\). Then, use the wave equation: \(v = f \lambda\), where \(v\) is speed and \(f\) is frequency. Rearranging for frequency gives: \(f = \frac{v}{\lambda} = \frac{0.24\text{ m/s}}{0.080\text{ m}} = 3.0\text{ Hz}\).

1 mark: Correct calculation of frequency by converting wavelength to metres and rearranging the wave equation.

First, calculate the maximum speed reached during acceleration: \(v = u + at = 0 + (2.0 \times 5.0) = 10\text{ m/s}\). Next, calculate the distance travelled during acceleration (the first stage): \(d_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 5.0\text{ s} \times 10\text{ m/s} = 25\text{ m}\). Then, calculate the distance travelled at a constant speed (the second stage): \(d_2 = \text{speed} \times \text{time} = 10\text{ m/s} \times 10\text{ s} = 100\text{ m}\). Total distance = \(d_1 + d_2 = 25\text{ m} + 100\text{ m} = 125\text{ m}\).

1 mark: Correct calculation of the total distance (125 m) by splitting the motion into acceleration and constant speed phases.

First, find the equivalent resistance of the parallel combination (\(R_p\)): \(\frac{1}{R_p} = \frac{1}{6.0} + \frac{1}{3.0} = \frac{1}{2.0}\), so \(R_p = 2.0\\ \Omega\). Next, find the total resistance of the circuit (\(R_t\)) by adding the series resistor: \(R_t = R_p + 2.0\\ \Omega = 2.0\\ \Omega + 2.0\\ \Omega = 4.0\\ \Omega\). Finally, calculate the total current (\(I\)) using Ohm's Law: \(I = \frac{V}{R_t} = \frac{12\text{ V}}{4.0\\ \Omega} = 3.0\text{ A}\).

1 mark: Correct calculation of the total resistance as 4.0 ohms and subsequently finding the total current as 3.0 A.

The proton number (atomic number) is 17, which is the number of protons. The nucleon number (mass number) is 37, so the number of neutrons is \(37 - 17 = 20\). The ion has a \(1-\) charge, which means it has gained one electron compared to a neutral chlorine atom. A neutral atom has 17 electrons, so this ion has \(17 + 1 = 18\) electrons.

1 mark: Identify 17 protons, 20 neutrons, and 18 electrons for the chloride ion.

Goblet cells secrete mucus, which traps dust, dirt, and pathogens such as bacteria. Ciliated cells have tiny hair-like structures called cilia, which beat rhythmically to sweep the trapped mucus upwards and away from the lungs toward the throat to be swallowed.

1 mark: Correctly identify that goblet cells produce mucus and ciliated cells sweep mucus away from the lungs.

In the reaction \(\text{Fe}_2\text{O}_3 + 3\text{CO} \rightarrow 2\text{Fe} + 3\text{CO}_2\), iron(III) oxide (\(\text{Fe}_2\text{O}_3\)) loses oxygen to become iron (\(\text{Fe}\)), meaning it is reduced. Carbon monoxide (\(\text{CO}\)) gains oxygen to become carbon dioxide (\(\text{CO}_2\)), meaning it is oxidised. Thus, carbon monoxide is the reducing agent.

1 mark: Identify the reduction of iron(III) oxide by carbon monoxide as the redox reaction where CO acts as the reducing agent.

(a) Speed is defined as distance traveled per unit time (or speed = distance / time). (b) Distance = speed \\times time = \(6\text{ m/s} \times 20\text{ s} = 120\text{ m}\). (c) Kinetic energy formula: \(KE = \frac{1}{2} m v^2\). Calculation: \(KE = 0.5 \times 80\text{ kg} \times (6\text{ m/s})^2 = 40 \times 36 = 1440\text{ J}\). (d) The energy is stored as chemical potential energy in muscles. (e) Air resistance increases because as speed increases, the bicycle collides with more air particles per second with greater force.

Part (a) [1 mark]: distance divided by time / distance per unit time. Part (b) [2 marks]: 1 mark for correct working (6 * 20) and 1 mark for correct answer of 120 (accept 120 m). Part (c) [3 marks]: 1 mark for correct formula (KE = 1/2 m v^2), 1 mark for correct substitution (0.5 * 80 * 6^2), 1 mark for correct answer of 1440 and unit Joules / J. Part (d) [1 mark]: chemical energy / chemical potential energy. Part (e) [1.8 marks]: 0.8 mark for stating air resistance increases, 1.0 mark for explanation (more collisions with air particles per unit time).

(a) Group I elements are called the alkali metals. Reactivity increases down the group. (b)(i) The word equation is: lithium + water -> lithium hydroxide + hydrogen. (ii) Observations can include: bubbles of gas / effervescence, the lithium floating, or the lithium disappearing. (c)(i) The formula of chlorine gas is \(\text{Cl}_2\). It is a pale green gas at room temperature. (ii) Test: place damp blue litmus paper (or indicator paper) into the gas. Positive result: the paper is bleached (turns white).

Part (a) [2 marks]: 1 mark for 'alkali metals', 1 mark for 'reactivity increases'. Part (b)(i) [2 marks]: 1 mark for reactants (lithium + water), 1 mark for products (lithium hydroxide + hydrogen). (ii) [1 mark]: effervescence / bubbles / floats / moves on water surface / dissolves (accept any one). Part (c)(i) [2 marks]: 1 mark for formula Cl2, 1 mark for pale green (or yellow-green) gas. (ii) [1.8 marks]: 1 mark for using damp blue litmus paper (or universal indicator paper), 0.8 mark for bleaching / turning white.

(a)(i) A hydrocarbon is a compound that consists of hydrogen and carbon atoms only. (ii) Ethene has a double covalent bond between the two carbon atoms and single covalent bonds with four hydrogen atoms: H2C=CH2. (b)(i) Unsaturated means the molecule contains at least one carbon-carbon double bond (C=C). (ii) The test is adding aqueous bromine (bromine water). Ethene (unsaturated) will decolourise the bromine water from orange/brown to colourless. Ethane (saturated) will show no change (remains orange/brown). (c) Complete combustion word equation: ethene + oxygen -> carbon dioxide + water.

Part (a)(i) [2 marks]: 1 mark for containing carbon and hydrogen, 1 mark for the word 'only' / 'solely'. (ii) [2 marks]: 1 mark for drawing C=C double bond, 1 mark for drawing four single C-H bonds correctly distributed. Part (b)(i) [1 mark]: contains a double bond / carbon-carbon double bond. (ii) [2.8 marks]: 1.0 mark for using bromine water / aqueous bromine, 0.9 mark for ethene decolourising (orange to colourless), 0.9 mark for ethane remaining orange / no change. Part (c) [1 mark]: ethene + oxygen -> carbon dioxide + water.

(a) In a longitudinal wave, the oscillations/vibrations of particles are parallel to the direction of wave travel (energy transfer). An example is sound waves. (b)(i) Frequency = number of cycles / time = 20 / 5.0 = \(4.0\text{ Hz}\). (ii) Formula: \(v = f \lambda\). Calculation: \(v = 4.0\text{ Hz} \times 1.5\text{ m} = 6.0\text{ m/s}\). (c)(i) Hazards of UV include sunburn, premature aging of skin, skin cancer, or damage to eyes (cataracts). (ii) A common use of infrared radiation is in TV remote controls (or thermal imaging, radiant heaters). Remote controls use infrared because it is easily transmitted through air and detected by sensors, and is safe (non-ionising).

Part (a) [2 marks]: 1 mark for definition (vibrations parallel to direction of wave travel), 1 mark for example (sound wave / ultrasound / seismic P-waves). Part (b)(i) [2 marks]: 1 mark for value 4.0, 1 mark for unit Hz (hertz). (ii) [2 marks]: 1 mark for formula v = f * lambda or substitution (4.0 * 1.5), 1 mark for correct answer 6.0 m/s (accept 6 or 6 m/s). Part (c)(i) [1 mark]: skin cancer / cataracts / sunburn / cell mutation. (ii) [1.8 marks]: 1.0 mark for valid use (remote controls / thermal imaging / cooking / optical fibres), 0.8 mark for explanation of suitability (e.g. low energy/safe, easily absorbed by food, detected by IR cameras).

(a)(i) Filtration is used to separate the insoluble sand (residue) from the soluble copper(II) sulfate solution (filtrate). (ii) First, heat the copper(II) sulfate solution to evaporate some water until the crystallization point is reached (or to obtain a hot, saturated solution). Then, leave the solution to cool down so crystals can form. Finally, filter the crystals from the remaining liquid and dry them between sheets of filter paper or in a warm oven. (b) Add aqueous sodium hydroxide (or aqueous ammonia) to the solution. A light blue precipitate will form, confirming the presence of copper(II) ions. (c) Add dilute hydrochloric acid (or dilute nitric acid) to the solution, followed by aqueous barium chloride (or aqueous barium nitrate). A white precipitate of barium sulfate will form.

Part (a)(i) [1 mark]: filtration. (ii) [3 marks]: 1 mark for heating/evaporating to obtain a saturated solution, 1 mark for cooling to allow crystals to form, 1 mark for filtering and drying the crystals (with filter paper / warm oven). Part (b) [2.8 marks]: 1.0 mark for adding aqueous sodium hydroxide (or ammonia), 1.8 marks for observing a blue precipitate (accept light blue ppt). Part (c) [2 marks]: 1 mark for adding acid (dilute nitric or hydrochloric acid) followed by barium solution (barium nitrate or barium chloride), 1 mark for observing a white precipitate.

(a)(i) A good dietary source of Vitamin C is citrus fruits (e.g., oranges, lemons) or strawberries. Its primary function is to maintain healthy skin, gums, blood vessels, and tissues (by helping to produce collagen). (ii) The deficiency disease is scurvy. (b)(i) Mechanical digestion begins in the mouth (buccal cavity), where teeth chew and grind food. (ii) Bile emulsifies fats, which means it breaks large fat globules into many smaller droplets. This increases the surface area of the fats, allowing the enzyme lipase to break them down much more rapidly. (Bile also neutralises stomach acid to create an alkaline pH for enzymes). (c) Chemical digestion is important because large, insoluble food molecules cannot be absorbed into the blood. Enzymes break them down into small, soluble molecules that can pass through the walls of the small intestine into the bloodstream.

Part (a)(i) [2 marks]: 1 mark for dietary source (citrus fruits / oranges / lemons / named fruit/vegetable rich in Vitamin C), 1 mark for function (making collagen / healthy skin / healthy gums / tissue repair). (ii) [1 mark]: scurvy. Part (b)(i) [1 mark]: mouth / teeth. (ii) [3 marks]: 1 mark for mentioning emulsification, 1 mark for explaining this breaks fats into smaller droplets to increase surface area, 1 mark for stating this allows faster digestion by lipase (or 1 mark for mentioning neutralisation of stomach acid / providing alkaline conditions). Part (c) [1.8 marks]: 1.0 mark for stating it breaks large insoluble molecules into small soluble molecules, 0.8 mark for explaining this allows absorption through the wall of the small intestine / into blood.

(a)(i) The SI unit of electric current is the ampere (A). (ii) A voltmeter must be connected in parallel with the resistor. (b)(i) The formula is \(V = I \times R\) (or \(R = V / I\)). Substituting the values: \(R = 6.0\text{ V} / 1.5\text{ A} = 4.0\text{ ohms}\) (or symbol \(\Omega\)). (ii) The formula for electrical power is \(P = V \times I\). Substituting the values: \(P = 6.0\text{ V} \times 1.5\text{ A} = 9.0\text{ watts}\) (W). (c) When a resistor is added in parallel, the total resistance of the circuit decreases (it becomes half of the original resistance, i.e., 2.0 ohms).

Part (a)(i) [1 mark]: ampere / amp / A. (ii) [1 mark]: parallel. Part (b)(i) [3 marks]: 1 mark for formula R = V / I (or V = I * R), 1 mark for calculation (6.0 / 1.5 = 4.0), 1 mark for unit ohms / Omega. (ii) [3 marks]: 1 mark for formula P = V * I (or P = I^2 * R), 1 mark for calculation (6.0 * 1.5 = 9.0), 1 mark for unit watts / W. Part (c) [0.8 mark]: decreases / is halved.

(a) A producer is an organism that makes its own organic nutrients (food), usually using energy from sunlight through photosynthesis. (b)(i) In this food chain, the herbivore is the rabbit, and the primary consumer is also the rabbit. (ii) The ultimate source of energy is sunlight (or the Sun). (iii) Energy is lost at each trophic level (about 90% is lost). Energy is lost through respiration, heat loss, movement, excretion, and because some parts of organisms are not eaten or digested. Therefore, after 4 or 5 levels, there is not enough energy remaining to support another trophic level. (c) Deforestation increases atmospheric CO2 because: 1. There are fewer trees carrying out photosynthesis, which absorbs CO2. 2. When forests are cleared, wood is often burned or left to decay, releasing stored carbon as CO2.

Part (a) [1 mark]: organism that makes its own organic nutrients/food (usually through photosynthesis / using light energy). Part (b)(i) [2 marks]: 1 mark for herbivore: rabbit, 1 mark for primary consumer: rabbit. (ii) [1 mark]: Sun / sunlight / light energy. (iii) [3 marks]: 1 mark for stating energy is lost at each stage / trophic level, 1 mark for giving examples of how energy is lost (respiration / heat / excretion / movement / uneaten parts), 1 mark for explaining that eventually too little energy is left to support another level. Part (c) [1.8 marks]: 1.0 mark for less photosynthesis to remove CO2, 0.8 mark for burning of wood/decomposition releasing CO2.

(a) Speed is calculated using the formula \(\text{speed} = \frac{\text{distance}}{\text{time}}\).
- Speed during the first 30 seconds: \(\frac{120\text{ m}}{30\text{ s}} = 4\text{ m/s}\).
- Speed during the last 10 seconds: \(\frac{180\text{ m}}{10\text{ s}} = 18\text{ m/s}\).
Therefore, the bicycle travels much faster in the last 10 seconds.

(b) Average speed is given by the formula:
\(\text{Average speed} = \frac{\text{total distance}}{\text{total time}}\)
- \(\text{Total distance} = 120\text{ m} + 180\text{ m} = 300\text{ m}\)
- \(\text{Total time} = 30\text{ s} + 20\text{ s} + 10\text{ s} = 60\text{ s}\)
- \(\text{Average speed} = \frac{300\text{ m}}{60\text{ s}} = 5\text{ m/s}\).

(c) (i) Weight is calculated using the formula:
\(W = m \times g\)
- \(W = 80\text{ kg} \times 10\text{ N/kg} = 800\text{ N}\).
(ii) Mass is a measure of the amount of matter in an object and does not change depending on location or gravitational field strength, so the mass remains unchanged (80 kg).

(a) [3 marks total]:
- [1 mark] Statement that the speed in the final section is faster / greater.
- [1 mark] Calculation of first speed: \(4\text{ m/s}\).
- [1 mark] Calculation of second speed: \(18\text{ m/s}\).

(b) [3 marks total]:
- [1 mark] Correct formula used or total distance determined as \(300\text{ m}\) and total time as \(60\text{ s}\).
- [1 mark] Correct calculation resulting in \(5\) or \(5.0\).
- [1 mark] Correct unit shown as \(\text{m/s}\) or \(\text{m s}^{-1}\).

(c)(i) [2 marks total]:
- [1 mark] Correct calculation of value: \(80 \times 10 = 800\).
- [1 mark] Correct unit: \(\text{N}\) (or Newtons).

(c)(ii) [1 mark total]:
- [1 mark] Correctly states that the mass remains the same / is unchanged.

(a) Hydrocarbon \( \text{X} \) (\( \text{C}_3\text{H}_6 \)) contains a carbon-to-carbon double bond and fits the general formula \( \text{C}_n\text{H}_{2n} \), which means it belongs to the alkenes homologous series. The displayed formula must show all atoms and all bonds, including a double bond between two carbon atoms: \( \text{H}_2\text{C}=\text{CH}-\text{CH}_3 \).

(b) (i) Bromine water is orange/brown. When added to an unsaturated hydrocarbon like propene, it is decolourised, changing from orange/brown to colourless.
(ii) This is an addition reaction where bromine atoms add across the double bond. The molecular formula of the product (1,2-dibromopropane) is \( \text{C}_3\text{H}_6\text{Br}_2 \).

(c) (i) During polymerization, the double bond opens up. The repeat unit must show two main-chain carbon atoms with single bonds extending out of the brackets, and the appropriate side groups: one carbon has two hydrogens, the other has one hydrogen and one methyl group (\( -\text{CH}_3 \)).
(ii) The polymer formed from propene monomers is poly(propene) (or polypropylene).

Total: 9 marks (scaled to 8.8)
(a) Alkene homologous series [1 mark]; Correct displayed formula showing all bonds, including C=C [1 mark] and all C-H and C-C bonds properly represented [1 mark].
(b) (i) Orange / brown / red-brown to colourless [1 mark] (Reject: clear).
(ii) Addition reaction [1 mark]; Correct formula \( \text{C}_3\text{H}_6\text{Br}_2 \) [1 mark].
(c) (i) Correct drawing of the repeat unit with single C-C bond, brackets, and open-ended bonds extending beyond brackets [1 mark]; correct substituent groups (\( -\text{H} \) and \( -\text{CH}_3 \)) attached to carbons [1 mark].
(ii) Poly(propene) or polypropylene [1 mark].

(a) (i) The combined resistance \( R_p \) is given by Ohm's Law: \( R_p = \frac{V}{I_t} = \frac{12.0\text{ V}}{3.0\text{ A}} = 4.0\ \Omega \).

(ii) For a parallel circuit: \( \frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} \). Substituting the known values: \( \frac{1}{4.0} = \frac{1}{6.0} + \frac{1}{R} \). Rearranging gives: \( \frac{1}{R} = \frac{1}{4.0} - \frac{1}{6.0} = \frac{3}{12} - \frac{2}{12} = \frac{1}{12} \). Therefore, \( R = 12.0\ \Omega \).

(iii) Since the resistors are in parallel, the potential difference across resistor \( \text{A} \) is equal to the battery voltage, \( 12.0\text{ V} \). The current is: \( I_A = \frac{V}{R_A} = \frac{12.0\text{ V}}{6.0\ \Omega} = 2.0\text{ A} \).

(b) The voltage across resistor \( \text{B} \) is also \( 12.0\text{ V} \). The power \( P \) dissipated by resistor \( \text{B} \) is: \( P = \frac{V^2}{R_B} = \frac{12.0^2}{12.0} = 12.0\text{ W} \) (or using \( I_B = 1.0\text{ A} \), \( P = I_B \times V = 1.0\text{ A} \times 12.0\text{ V} = 12.0\text{ W} \)).

Total: 9 marks (scaled to 8.8)
(a) (i) Formula \( R = \frac{V}{I} \) [1 mark]; correct calculation to give \( 4.0\ \Omega \) [1 mark].
(ii) Formula \( \frac{1}{R_p} = \frac{1}{R_A} + \frac{1}{R_B} \) [1 mark]; correct algebraic step showing \( \frac{1}{R} = \frac{1}{12} \) leading to \( R = 12\ \Omega \) [1 mark].
(iii) Statement or use of \( V = 12.0\text{ V} \) across parallel branch [1 mark]; correct calculation of \( I_A = 2.0\text{ A} \) [1 mark].
(b) Correct formula for power, e.g., \( P = \frac{V^2}{R} \) or \( P = I^2 R \) [1 mark]; correct calculation of power as \( 12.0 \) [1 mark]; correct unit of Watts or \( \text{W} \) [1 mark].

(a) Using Snell's Law: \( n = \frac{\sin(i)}{\sin(r)} \). Rearranging for \( \sin(r) \): \( \sin(r) = \frac{\sin(i)}{n} = \frac{\sin(45^\circ)}{1.50} = \frac{0.7071}{1.50} = 0.4714 \). Calculating the inverse sine: \( r = \sin^{-1}(0.4714) = 28.1^\circ \).

(b) The critical angle is the angle of incidence in the optically denser medium (glass) at which the angle of refraction in the less dense medium (air) is exactly \( 90^\circ \).

(c) Using the formula: \( \sin(c) = \frac{1}{n} \), where \( n = 1.50 \). Thus, \( \sin(c) = \frac{1}{1.50} = 0.6667 \). Calculating the inverse sine: \( c = \sin^{-1}(0.6667) = 41.8^\circ \).

(d) Regions of the electromagnetic spectrum with wavelengths shorter than visible light are ultraviolet (UV), X-rays, and gamma rays.

Total: 9 marks (scaled to 8.8)
(a) Formula \( n = \frac{\sin(i)}{\sin(r)} \) [1 mark]; substitution and rearrangement to find \( \sin(r) = 0.471 \) [1 mark]; final angle \( 28.1^\circ \) (accept \( 28^\circ \)) [1 mark].
(b) Defined as the angle of incidence in the denser medium [1 mark] that results in an angle of refraction of \( 90^\circ \) (or along the boundary) [1 mark].
(c) Formula \( \sin(c) = \frac{1}{n} \) [1 mark]; correct calculation of \( c = 41.8^\circ \) (accept \( 42^\circ \)) [1 mark].
(d) Any two correct regions from: ultraviolet, X-rays, gamma rays [2 marks, 1 mark for each].

(a) Acceleration \( a \) is calculated using: \( a = \frac{v - u}{t} = \frac{8.0\text{ m/s} - 0}{4.0\text{ s}} = 2.0\text{ m/s}^2 \).

(b) Total distance is the area under the speed-time graph. The motion consists of three phases:
1. Acceleration phase (triangle): \( \text{Area}_1 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4.0\text{ s} \times 8.0\text{ m/s} = 16.0\text{ m} \).
2. Constant speed phase (rectangle): \( \text{Area}_2 = \text{base} \times \text{height} = 6.0\text{ s} \times 8.0\text{ m/s} = 48.0\text{ m} \).
3. Deceleration phase (triangle): \( \text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2.0\text{ s} \times 8.0\text{ m/s} = 8.0\text{ m} \).
Total distance = \( 16.0 + 48.0 + 8.0 = 72.0\text{ m} \).

(c) Kinetic energy \( E_k \) is given by: \( E_k = \frac{1}{2} m v^2 = \frac{1}{2} \times 1.5\text{ kg} \times (8.0\text{ m/s})^2 = 0.75 \times 64 = 48.0\text{ J} \).

Total: 9 marks (scaled to 8.8)
(a) Formula \( a = \frac{\Delta v}{t} \) [1 mark]; correct substitution and calculation \( 2.0 \) [1 mark]; correct unit \( \text{m/s}^2 \) [1 mark].
(b) Idea that distance is area under speed-time graph [1 mark]; correct calculation of individual component areas (e.g., 16, 48, 8) [1 mark]; correct total sum of \( 72.0\text{ m} \) [1 mark].
(c) Formula \( E_k = \frac{1}{2} m v^2 \) [1 mark]; substitution of values: \( 0.5 \times 1.5 \times 8.0^2 \) [1 mark]; final answer with correct unit: \( 48.0\text{ J} \) [1 mark].

(a) Place the calcium carbonate chips in a conical flask, add dilute hydrochloric acid, and immediately seal with a stopper connected to a gas syringe. Measure the volume of carbon dioxide gas collected in the syringe at regular, specified time intervals (e.g., every 10 seconds).

(b) (i) When the temperature increases, the particles gain kinetic energy and move faster. This results in more frequent collisions per unit time. Additionally, a greater proportion of the colliding particles possess energy equal to or greater than the activation energy, leading to a higher percentage of successful collisions.

(ii) At a higher temperature, the initial gradient (slope) of the curve is steeper, representing a faster rate of reaction. The curve will level off sooner, but will reach the same final volume of carbon dioxide because the quantity of limiting reactant (hydrochloric acid) remains unchanged.

Total: 9 marks (scaled to 8.8)
(a) Suitable apparatus specified: gas syringe/conical flask (or balance and flask for mass loss) [1 mark]; measurement of dependent variable: volume of gas (or mass loss) [1 mark]; measurement over regular time intervals [1 mark].
(b) (i) Particles gain kinetic energy / move faster [1 mark]; more frequent collisions / higher collision frequency [1 mark]; more particles have energy \( \ge \text{activation energy} \) [1 mark]; higher frequency of successful collisions [1 mark].
(ii) Graph has a steeper initial slope/gradient [1 mark]; curve plateaus/levels off earlier at the same final volume of gas [1 mark].

(a) Reactivity increases down Group I. As you go down the group, the atomic radius increases and there are more electron shells, meaning the single outer shell electron is further from the positively charged nucleus. This results in weaker electrostatic attraction between the nucleus and the outer electron, so the electron is lost more easily.

(b) (i) At room temperature and pressure, chlorine is a pale green gas.
(ii) Chlorine is more reactive than iodine, so it displaces iodine from potassium iodide. The balanced equation with state symbols is: \( \text{Cl}_2(\text{g}) + 2\text{KI}(\text{aq}) \rightarrow 2\text{KCl}(\text{aq}) + \text{I}_2(\text{aq}) \).
(iii) Chlorine atoms gain electrons to form chloride ions (reduction: \( \text{Cl}_2 + 2\text{e}^- \rightarrow 2\text{Cl}^- \)), while iodide ions lose electrons to form iodine molecules (oxidation: \( 2\text{I}^- \rightarrow \text{I}_2 + 2\text{e}^- \)). Since both electron loss and gain occur, it is a redox reaction.

Total: 9 marks (scaled to 8.8)
(a) Reactivity increases down Group I [1 mark]; outer shell electron is further from the nucleus / more shielded [1 mark]; less attraction from the nucleus makes it easier to lose the outer electron [1 mark].
(b) (i) Pale green [1 mark]; gas [1 mark].
(ii) Correct chemical formulae of reactants and products (\( \text{Cl}_2 \), \( \text{KI} \), \( \text{KCl} \), \( \text{I}_2 \)) [1 mark]; correctly balanced [1 mark]; correct state symbols: \( (\text{g}) \) and \( (\text{aq}) \) [1 mark].
(iii) Oxidation is loss of electrons (by iodide) and reduction is gain of electrons (by chlorine) [1 mark].

(a) (i) Palisade mesophyll cells are located near the upper surface of the leaf to absorb maximum sunlight. They are packed with a large number of chloroplasts containing chlorophyll and are arranged vertically to capture light efficiently.
(ii) Guard cells control the opening and closing of stomata. They have cell walls of uneven thickness (thicker on the inner side), which allows them to bend and open the stomatal pore when turgid (water enters), facilitating gas exchange, and close it when flaccid to prevent water loss.

(b) (i) The student can vary the light intensity by changing the distance between the light source (lamp) and the beaker containing the aquatic plant.
(ii) The dependent variable can be measured by counting the number of oxygen bubbles released per minute, or by measuring the volume of oxygen gas collected in a gas syringe over a fixed time period.

(c) As light intensity increases, the rate of photosynthesis eventually plateaus because another factor becomes limiting. This limiting factor is typically carbon dioxide concentration or temperature. At very high light intensities, the enzymes involved in photosynthesis are working at their maximum possible rate (saturated), and cannot process substrates any faster unless temperature or carbon dioxide levels are increased.

Total: 9 marks (scaled to 8.8)
(a) (i) Located near upper surface/vertically arranged [1 mark]; contain many chloroplasts to absorb maximum light [1 mark].
(ii) Positioned around stomata to control opening and closing [1 mark]; wall thickness variations allow shape change based on turgor pressure [1 mark].
(b) (i) Move the lamp to different distances from the plant [1 mark].
(ii) Number of bubbles per unit time / volume of oxygen gas collected [1 mark].
(c) Reference to limiting factors [1 mark]; carbon dioxide concentration or temperature is limiting [1 mark]; enzymes working at maximum capacity / active sites saturated [1 mark].

(a) According to the lock-and-key hypothesis, salivary amylase has a specifically shaped region called the active site, which is complementary only to the shape of the starch substrate (the key). Starch binds to the active site of amylase to form an enzyme-substrate complex, where it is broken down. Proteins have a different 3D shape and do not fit into the active site of amylase, so no reaction occurs.

(b) (i) A very low pH of 2 is highly acidic. This extreme pH breaks the bonds holding the 3D structure of the amylase protein together, causing it to denature. Denaturation permanently alters the shape of the enzyme's active site, meaning starch molecules can no longer fit into it, and enzymatic activity is lost.
(ii) To test for the reducing sugar maltose, Benedict's test is used. When heated with Benedict's solution, the mixture changes from blue to green, yellow, orange, or brick-red, confirming the presence of reducing sugars.

Total: 9 marks (scaled to 8.8)
(a) Active site has a specific 3D shape [1 mark]; shape of active site is complementary to starch (the substrate) [1 mark]; forms enzyme-substrate complex [1 mark]; protein substrate does not fit because its shape is not complementary to the active site [1 mark].
(b) (i) Low pH (highly acidic conditions) denatures the enzyme [1 mark]; changes/destroys the shape of the active site [1 mark]; substrate can no longer bind / fit [1 mark].
(ii) Benedict's test [1 mark]; heat and observe brick-red precipitate / colour change from blue to green/yellow/orange/red [1 mark].

(a) When the concentration of hydrochloric acid increases, the rate of reaction increases. This is because there are more reactant particles per unit volume, which increases the frequency of collisions (more collisions occurring per unit time) between the reacting particles.

(b) (i) The initial rate of production of carbon dioxide gas decreases.
(ii) One large piece of marble has a smaller total surface area than small chips of the same mass. This means fewer calcium carbonate particles are exposed to the acid, resulting in a lower frequency of collisions per unit time.

(c) The balanced chemical equation with state symbols is:
\(\text{Mg}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{MgCl}_2(\text{aq}) + \text{H}_2(\text{g})\)

(a) [Total: 3.0 marks]
- Rate of reaction increases [1 mark]
- More acid particles per unit volume [1 mark]
- Higher frequency of collisions / more collisions per unit time [1 mark]

(b)(i) [Total: 1.0 mark]
- Rate decreases / reaction is slower [1 mark]

(b)(ii) [Total: 2.0 marks]
- Smaller total surface area [1 mark]
- Lower collision frequency / fewer collisions per unit time [1 mark]

(c) [Total: 2.8 marks]
- Correct chemical formulae for reactants and products: \(\text{Mg}\), \(\text{HCl}\), \(\text{MgCl}_2\), \(\text{H}_2\) [1.0 mark]
- Correct balancing: \(\text{Mg} + 2\text{HCl} \rightarrow \text{MgCl}_2 + \text{H}_2\) [1.0 mark]
- All state symbols correct: \(\text{(s)}\), \(\text{(aq)}\), \(\text{(aq)}\), \(\text{(g)}\) [0.8 marks]

(a) Using the formula rate = 1000 / time:
- For pH 4: 1000 / 300 = 3.3
- For pH 5: 1000 / 180 = 5.6
- For pH 6: 1000 / 60 = 16.7
- For pH 7: 1000 / 120 = 8.3
- For pH 8: 1000 / 240 = 4.2
(b) The independent variable (what is changed) is the pH. The dependent variable (what is measured) is the time taken for starch to disappear, or the calculated rate of reaction.
(c) Key control variables include keeping the reaction temperature constant (e.g. using a water bath), maintaining the same concentrations and volumes of starch and amylase solutions.
(d) Iodine solution is a yellow-brown indicator that turns blue-black in the presence of starch. If starch is fully broken down, the iodine will remain yellow-brown.
(e) At extreme pH values like pH 4, the bonds holding the amylase protein structure are disrupted, denaturing the enzyme. The active site changes shape, meaning starch molecules can no longer fit.
(f) A suitable control replaces the active enzyme with an equal volume of distilled water (or inactive, boiled amylase) to show that starch breakdown does not occur without active amylase.

Part (a): 2 marks total. 1 mark for at least 3 correct rates, 2 marks for all 5 correct: 3.3, 5.6, 16.7, 8.3, 4.2 (allow rounding to 1 d.p.).
Part (b): 2 marks total. 1 mark for identifying pH as independent, 1 mark for identifying rate or time taken as dependent.
Part (c): 2 marks total. 1 mark for each valid constant variable (accept: temperature, volume of starch, concentration of starch, volume of amylase, concentration of amylase).
Part (d): 1 mark for stating that the iodine remains orange-brown / does not turn blue-black.
Part (e): 2 marks total. 1 mark for stating the enzyme is denatured / active site changed shape, 1 mark for explaining that the substrate (starch) can no longer fit or bind.
Part (f): 1 mark for stating that amylase should be replaced with distilled water (or boiled amylase).

(a) The rate of reaction is initially very high, shown by the rapid increase in gas volume (18 cm3 in the first 30 s). The rate then slows down as time progresses, with smaller increases in volume (e.g., only 2 cm3 between 150 s and 180 s). By 210 s, the rate is zero as the volume remains constant at 54 cm3.
(b) As the reaction proceeds, reactant particles (hydrochloric acid ions) are consumed. The concentration of acid decreases, which reduces the frequency of successful collisions between reactant particles.
(c) The constant reading indicates that the reaction has reached completion because at least one reactant (likely the limiting reactant) is fully exhausted.
(d) Fine powder has a larger surface area than chips, which increases the rate of reaction. The curve on the graph will be steeper initially but will level off to the exact same volume of 54 cm3, reaching this plateau at an earlier time.
(e) Dilute hydrochloric acid is an irritant, so safety goggles must be worn to prevent splashes into the eyes. (Accept: lab coat/gloves).
(f) The standard chemical test for carbon dioxide is bubbling the gas through limewater, which turns from clear to cloudy/milky.

Part (a): 3 marks total. 1 mark for stating that the rate is fastest at the beginning, 1 mark for stating that the rate slows down over time, 1 mark for stating that the reaction stops/plateaus at 210 s with 54 cm3.
Part (b): 2 marks total. 1 mark for stating that reactant concentration decreases / particles are used up, 1 mark for fewer successful collisions per unit time.
Part (c): 1 mark for stating that the reaction has finished or a reactant is used up.
Part (d): 2 marks total. 1 mark for stating the initial curve is steeper, 1 mark for stating it reaches the same final volume of 54 cm3 / plateau at an earlier time.
Part (e): 1 mark for safety goggles (accept gloves/lab coat).
Part (f): 1 mark for limewater turning cloudy/milky.

(a) The ammeter must be connected in series with the wire under test to measure the current passing through it. The voltmeter must be connected in parallel across the two ends of the test wire to measure the potential difference across it.
(b) Calculating resistance R = V / I:
- 20.0 cm: 0.60 / 0.50 = 1.2 Ohm
- 40.0 cm: 1.20 / 0.50 = 2.4 Ohm
- 60.0 cm: 1.80 / 0.50 = 3.6 Ohm
- 80.0 cm: 2.40 / 0.50 = 4.8 Ohm
- 100.0 cm: 3.00 / 0.50 = 6.0 Ohm
The unit of resistance is the Ohm (represented by the symbol Omega).
(c) The resistance is directly proportional to the length of the wire. This is supported by the data: when the length doubles from 20.0 cm to 40.0 cm, the resistance doubles from 1.2 Ohm to 2.4 Ohm (or when length increases 5-fold from 20 to 100 cm, resistance increases 5-fold from 1.2 to 6.0 Ohm).
(d) The temperature of the wire must be kept constant. This is controlled by opening the switch to turn off the current between readings to prevent heating, or by keeping the current very low. (Alternatively, thickness/cross-sectional area of the wire, controlled by using the exact same piece of wire throughout the experiment).
(e) When current flows through a wire, it generates thermal energy (heats up). An increase in temperature increases the vibration of the lattice ions in the metal, which increases resistance. Keeping the switch closed would cause systematic errors by progressively increasing the resistance readings.

Part (a): 2 marks total. 1 mark for ammeter in series, 1 mark for voltmeter in parallel across the test wire.
Part (b): 2 marks total. 1 mark for calculating all resistance values correctly (1.2, 2.4, 3.6, 4.8, 6.0), 1 mark for stating the unit is Ohm (or symbol Omega).
Part (c): 2 marks total. 1 mark for stating the relationship is directly proportional / linear, 1 mark for quantitative justification using two data points.
Part (d): 2 marks total. 1 mark for identifying a correct variable (temperature or wire diameter), 1 mark for describing how to control it (open switch between readings or use the same wire/gauge).
Part (e): 2 marks total. 1 mark for stating that current heats the wire, 1 mark for explaining that higher temperature increases resistance / makes the test unfair.

(a) Calculating the temperature drops:
- Beaker A (matte black): 90.0 °C - 65.0 °C = 25.0 °C
- Beaker B (shiny silver): 90.0 °C - 78.5 °C = 11.5 °C
(b) Beaker A lost thermal energy at a faster rate. This is because matte black surfaces are much better emitters of infrared/thermal radiation than shiny, light-colored surfaces (which are poor emitters).
(c) Variables to keep constant:
1. The starting temperature of the water (90.0 °C in both).
2. The volume of water in each beaker (150 cm3).
3. The surrounding room temperature and environmental drafts.
4. The thickness/material of the beakers.
(d) Placing an insulated lid (e.g., cardboard or plastic) on top of each beaker. This reduces thermal energy transfer via convection and evaporation from the surface of the water, making sure that any observed temperature differences are due to the differences in radiation from the vertical sides of the beakers.
(e) A measuring cylinder is the standard laboratory tool used to measure the volume of liquid (150 cm3).

Part (a): 2 marks total. 1 mark for correct calculation for Beaker A (25.0 °C), 1 mark for correct calculation for Beaker B (11.5 °C). (Do not penalize missing units).
Part (b): 2 marks total. 1 mark for identifying Beaker A, 1 mark for explaining that black surfaces are better/more efficient emitters of thermal radiation than shiny silver surfaces.
Part (c): 3 marks total. 1 mark for each valid control variable listed (accept: volume of water, initial temperature, same type of beaker, same room temperature / no drafts), up to a maximum of 3 marks.
Part (d): 2 marks total. 1 mark for suggesting a lid / stopper, 1 mark for explaining that it stops/minimizes evaporation or convection from the top of the liquid.
Part (e): 1 mark for specifying a measuring cylinder.